# Jamb mathematics tutorial: sequences and series

Hi guys. Today we’ll be looking to completely exhaust **sequences and series** based on **jamb mathematics syllabus**.

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## 1.1 Sequences and series

A set of numbers such as U_{1}, U_{2}, U_{3}, U_{4}, \ldots, U_{n} which are ordered is called a sequence. U_{1} is the first term, U_{2} is the second term, U_{n} is the n_{th} term of the sequence. The following are examples of sequences

**1. **1,3,5,7,\ldots,2n-1 (sequence of odd numbers)

**2. **2,3,6,8,10,\ldots,2n (sequence of even numbers)

The expression U_{1}+U_{2}+U_{3}+U_{4}+\ldots+U_{n} which can be formed from a sequence is called a series.

## 1.2 Arithmetic progression (A.P)

An arithmetic progression of A.P is a series of the form: a+(a+d)+(a+2d)+(a+3d)+\ldots where a is the first term and d is called the common difference which may be positive or negative. Each bracket of the A.P is a term and simple progression will give the n_{th} term of the sequence.

U_{n}=a+(n-1)dIf there are n terms in the A.P and S_{n} is their sum, then

S_{n}=[a]+[a+d]+[a+2d]+\ldots+[a+(n-1)d]Reversing the series gives

S_{n}=[a+(n-1)d]+[a+(n-2)d]+[a+(n-3)d]+\ldots+[a]Adding both equations of S_{n}gives

2S_{n}=[2a+(n-1)d]+[2a+(n-1)d]+[2a+(n-1)d]+\ldots+[2a+(n-1)d] \\ 2S_{n}=n[2a+(n-1)d] \\ S_{n}=\frac{n}{2}[2a+(n-1)d]{Sum of n terms of a sequence}

U_{1}= first term =a, U_{n}= last term =a+(n-1)d\\\therefore S_{n}=\frac{n}{2}[U_{1}+U_{n}] {Sum of n terms of a sequence}

## 1.3 Arithmetic mean

If p,q,r are three consecutive terms of an A.P, then q is called the arithmetic mean of p and r. Since p,q,r are consecutive terms of an A.P, it implies that

q-p=r-q=d \\ q-p=r-q\\2q=r+p\\q=\frac{r+p}{2}The arithmetic mean of two numbers is the average of the two numbers.

**Example 1**

**a.** Find the 15_{th} term and the sum of the first 26 terms of the series 21,17,13,9,5,\ldots

**b. **Calculate the arithmetic mean of -15 and -23

**Solution**

Given 21,17,13,9,5,\ldots \\ a=21, d=17-21=-4, n=15, U_{15}=?, S_{26}=?

**a. i. **U_{n}=a+(n-1)d \\U_{15}=21+(15-1)(-4) \\U_{15}=21+(14)(-4)\\U_{15}=21-56\\U_{15}=-35

**ii. **S_{n}=\frac{n}{2}[2a+(n-1)d]\\S_{26}=\frac{26}{2}[2(21)+(26-1)(-4)]\\S_{26}=13[42+(25)(-4)]\\S_{26}=13[42-100]\\S_{26}=13[-58]\\S_{26}=-754

alternatively,

U_{26}=21+25(-4)\\U_{26}=21-100\\U_{26}=-79\\S_{n}=\frac{n}{2}[U_{1}+U_{n}]\\S_{26}=\frac{26}{2}[21+(-79)]\\S_{26}=13[-58]\\S_{26}=-754**b.** arithmetic mean =\frac{-15+(-23)}{2}=-19

## 1.4 Geometric progression (G.P)

A series that is in the form a+ar+ar^2+ar^3+\ldots is called a geometric progression, where a is the first term and r is the common ratio. The n_(th) term is U_{n}=ar^{n-1}.

Let S_{n} be the sum of the n_{th} terms of a series

S_{n}= a+ar+ar^2+ar^3+\ldots+ar^{n-1}multiply through by r

rS_{n}=ar+ar^2+ar^3+\ldots+ar^{n-1}+ar^{n}subtracting $rS_{n}$ from $S_{n}[/latex] gives

(1-r)S-{n}=a-ar^n\\S_{n}=\frac{a(1-r^n)}{1-r} (first equation)

subtracting S_{n} from rS_{n} gives

(r-1)S_{n}=ar^n-a\\S_{n}=\frac{a(r^n-1)}{r-1} (second equation)

Both equations can be used to solve for S_{n}

## 1.5 Geometric mean

If the number a,b,c are three consecutive terms of a G.P, then b is called the geometric mean of a and c. Since a,b,c are consecutive terms of a G.P,

\frac{b}{a}=\frac{c}{b}=r\\b^2=ac\\b=\sqrt{ac}The value of b could either be +ve or -ve but never both. The common ratio determines it.

**Example 2**

**a. **Find the sum of the first 8 terms of the series \frac{1}{27}, \frac{1}{9}, \frac{1}{3},\ldots

**b. **If the 7_{th} and 10_{th} terms of a G.P are 320 and 2560 respectively, find the series

**Solution**

**a. **Given \frac{1}{27}, \frac{1}{9}, \frac{1}{3},\ldots\\a=\frac{1}{27}, r=\frac{1}{9}\div \frac{1}{27}=3, n=8\\S_{n}=\frac{a(1-r^n)}{1-r}\\S_{8}=\frac{\frac{1}{27}(1-3^8)}{1-3}= \frac{-6560}{27\times-2}\\S_{8}=\frac{-6560}{-54}\\S_{8}=121.48

alternatively,

S_{n}=\frac{a(r^n-1)}{r-1}\\S_{8}=\frac{\frac{1}{27}(3^8-1)}{3-1}= \frac{6560}{27\times2}\\S_{8}=\frac{6560}{54}\\S_{8}=121.48It’s obvious that both equations can work

**b. **7_{th} term =ar^{7-1}=ar^6\\ar^6=320

Similarly, 10_{th} term =ar^9\\ar^9=2560\\\frac{ar^9}{ar^6}=\frac{2560}{320}\\r^3=8\\r=\sqrt[3]{8}\\r=2\\ar^6=320\\a(2)^6=320\\a=\frac{320}{64}\\a=5

Remember a geometric series is represented as a,ar,ar^2,ar^3,\ldots\\\therefore the G.P is 5,10,20,40

### 1.6 Sum to infinity of a G.P

S_{\infty}=\frac{a}{1-r}Where /r/<1 i.e the magnitude of r whether positive or negative is always less than 1.

\therefore ranges from -1 to + 1 i.e -1<r<1 but r\ne0

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Abubakar says

Solve: The second and fifth term of a GP are 3/2 and 1/12, respectively. What is the first term?

Abubakar says

Ok thanks. Please note that this question is a JAMB past question but the year of release is not specified.